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Math Help - piecewise functions and continuity.

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    piecewise functions and continuity.



    I know the answer to 1a is No, but i dont know why, just wanting to know the theory behind it and i dont know how to do 1b...any help would be appreciated. thanks
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    Quote Originally Posted by b00yeah05 View Post


    I know the answer to 1a is No, but i dont know why, just wanting to know the theory behind it and i dont know how to do 1b...any help would be appreciated. thanks
    each piece of the function is a polynomial, so we know each piece is continuous on its own. however, we must be concerned with where the function breaks. this is at x = 3.

    for 1a, k = 1 does not work because when we plug in k = 1 and x = 3 we end up with two different values for each piece, which obviously (why is this obvious?) makes the function not continuous.

    for 1b. we need to find k so that the function is continuous. meaning, when x = 3, we want both functions to give the same value. how do you think we should proceed?
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    well the obvious reason would be that it is different values and hence it cant be continuous for 1a, and as for 1b i was thinking maybe solve them as a simultaneous equation?
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    Quote Originally Posted by b00yeah05 View Post
    [snip]
    as for 1b i was thinking maybe solve them as a simultaneous equation?
    Not quite ....

    3^2 + 3k - 27 = -(3)^2 + 7(3) + k (and how did I get this?)

    => .......
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    you made x=3 and then solving...hence k = 24. at that value the function is continuous. am i correct??
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    Quote Originally Posted by b00yeah05 View Post
    you made x=3 and then solving...
    yes. but the important step is equating the two formulas. this ensures that they are equal when we're done
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    i edited my post before..sorry..am i correcT?
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    Quote Originally Posted by b00yeah05 View Post
    i edited my post before..sorry..am i correcT?
    no
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    what do i do after i get the co-efficient of k = 29?
    Last edited by b00yeah05; January 13th 2008 at 11:51 PM. Reason: changed value of k
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    Quote Originally Posted by b00yeah05 View Post
    what do i do after i get the co-efficient of k = 29?
    that is wrong as well. remember, we are solving 9 + 3k - 37 = -9 + 21 + k for k
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    ah...i think you might have missed the negative sign there...it is negative x^2 in the second function so (-3)^2 = 9 which would cancel the 9's out?
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    Quote Originally Posted by b00yeah05 View Post
    ah...i think you might have missed the negative sign there...it is negative x^2 in the second function so (-3)^2 = 9 which would cancel the 9's out?
    you have -x^2, so if you plug in x = 3 you have -(3)^2 = -9
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    i see...so thats where i was going wrong..so k=20
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    Quote Originally Posted by b00yeah05 View Post
    i see...so thats where i was going wrong..so k=20
    yes
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    Quote Originally Posted by b00yeah05
    ah...i think you might have missed the negative sign there...it is negative x^2 in the second function so (-3)^2 = 9 which would cancel the 9's out?
    Quote Originally Posted by Jhevon View Post
    you have -x^2, so if you plug in x = 3 you have -(3)^2 = -9
    *Ahem* b00yeah05 ........ which is why I wrote -(3)^2, NOT (-3)^2:

    Quote Originally Posted by mr fantastic View Post
    Not quite ....

    3^2 + 3k - 27 = -(3)^2 + 7(3) + k (and how did I get this?)

    => .......
    A flagon of frustration might have been saved if the original equation (above) that started this almost-death by a thousand cuts had been read and worked a little more ...... carefully?

    No need to thank me for the above (apparently useless) post - the pleasure's in the giving.
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